## How much is a vessel's performance influenced by wind?

(This is not scientific article as it only contains some assumptions to get an idea about the subject…)

First published on LinkedIn on 24th July 2017, by Alexander Schepers. Please check for some really interesting comments and nice links within the discussion.

I was always wondering by how much a vessel is influenced ** by wind** in comparison to the force needed to propel her. It was to be found a simple formula about the drag factor (cw) of the wind exposed area of the accomodation + wave braker and (to keep it simple) that of a more beneficial body such as a sphere.

Wikipedia was my friend and it gave: **Force** = (**density of air** x **cw** x **area** x **speed of wind hitting exposed areas**(*square*) ) / **2**

The area exposed to wind is the accomodation and the wave braker in the bow of the ship (drag factor 1,17). The idea is to “shield” the accomodation and wave braker with something like a sphere (drag factor 0,34) (source: __http://www.uni-magdeburg.de/isut/LSS/Lehre/Arbeitsheft/VII.pdf____)__

Let’s imagine our ship (a general cargo vessel of 8.000 dwt) has a relatively high accomodation of 11 meters and a width of 12 meters and a wave braker of 14 x 6 meters. It sails at a speed 13 knots ( = 6,69 m/s) in straight head winds of beaufort 6 (= 12.35 m/s), so you need to add the windspeed onto the vessel’s speed (=19,04 m/s).

So we have (1,2041 (kg/m3) x 1,17 x (11m x 12m + 14m x 6m) x (19,04) (m/s) x (19,04) (m/s) / 2 = 55.157 (kg x m4) / (m3 x s2) = 55.157 kg m / s2 = ~55 kN

If you replace the drag factor in above calculation with that of the sphere (0,34), and the enlarged area of the sphere (0,5 x pi x diameter = 0,5 x 3,142 x 12 x 12 = 226 m2 and assume for the wave breaker 130 m2) then you have a force of only 26 kN.

Hence difference is some 29 kN.

The ultimate question is, how much force is needed to propel the vessel? This information you will usually find in the towing tank test data of the vessel. If you do not know your way around, look for data labelled with the unit of force: kN and read backwards into the paper. It is dependent on speed.

For our example it is 300 kN at this speed.

Here you have it – you would potentially

if you would shield the accomodation and the wave braker with a sphere.save 10% energy

The potential is relatively high, but here are some thoughts:

- You need to know prevailing winds (direction and speed) of your trade and the trade’s speed throughout the year to make a good estimation of the potential!
- You need to offset the wind direction and make an assumption how it would influence the force (i.e. if you have winds from direction 45° of vessel’s course, would you need to reduce the wind speed hitting the exposed area by 50%; 90° = 0% (only vessel’s speed); 135° = pushing winds with 50% of wind speed; 180° = 100% pushing; 225° = 50% pushing; 270° = 0%; 315° = adding 50%). That means you need to have data available with average wind directions, speed and time.
- Aerodynamics are influenced by a myriad of factors and likely predominatly the cargo cranes and bow section.
- How do you want to build a sphere or any other object shielding the most wind exposed areas? Are building costs exceeding the CAPEX?
- Materials, handling times to build the sphere up and disassembly when you need the space.

If you need more information or guidance as to how to collect data and compile it with Excel to make an appropriate study, just let me know.